The following data are obtained when dinitrogen and dioxygen react together to form different compounds: | | Mass of dinitrogen | Mass of dioxygen | |---|---|---| | (i) | 14 g | 16 g | | (ii) | 14 g | 32 g | | (iii) | 28 g | 32 g | | (iv) | 28 g | 80 g | (a) Which law of chemical combination is obeyed by the above experimental data? Give its statement. (b) Fill in the blanks in the following conversions: (i) 1 km = ............ mm = ............ pm (ii) 1 mg = ............ kg = ............ ng (iii) 1 mL = ............ L = ............ $dm^3$

The following data are obtained when dinitrogen an – Class 11 Chemistry

Exercises · 1.21

The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

	Mass of dinitrogen	Mass of dioxygen
(i)	14 g	16 g
(ii)	14 g	32 g
(iii)	28 g	32 g
(iv)	28 g	80 g

(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement. (b) Fill in the blanks in the following conversions: (i) 1 km = ............ mm = ............ pm (ii) 1 mg = ............ kg = ............ ng (iii) 1 mL = ............ L = ............ $dm^3$

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Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

Imagine you're measuring the length of your desk. You pull out a ruler marked in centimetres and find it's 120 cm long. Your friend, using a metre stick, says it's 1.2 m. You're both right — you've just used different units to describe the same physical length.

That's the core idea: unit conversion is the process of changing how you express a quantity without changing the quantity itself.

The Intuition: Same Quantity, Different Labels

Think of a pizza. Whether you call it "one pizza" or "8 slices," the amount of pizza hasn't changed. You've just used a different unit (pizza vs. slice) to describe it.

Similarly, 1 metre and 100 centimetres are the same length — just like 1 pizza and 8 slices are the same amount. The number changes (1 becomes 100, or 1 becomes 8), but the actual thing being measured stays identical.

Note

This is the most important idea to hold onto: conversion changes the number, not the quantity. If you ever feel like the quantity has changed, you've made a mistake.

The Precise Statement

Unit conversion is the multiplication of a quantity by a conversion factor — a fraction equal to 1 — that cancels the old unit and introduces the new one.

A conversion factor looks like this:

\frac{\text{new unit}}{\text{old unit}} = 1

For length: $\frac{1 \text{ m}}{100 \text{ cm}} = 1$ and $\frac{100 \text{ cm}}{1 \text{ m}} = 1$ .

Why are these fractions equal to 1? Because 1 metre is 100 centimetres. The numerator and denominator describe the same physical length, so their ratio is exactly 1.

How to Convert: The Only Rule You Need

Multiply by a conversion factor that cancels the unit you have and leaves the unit you want.

Let's convert 120 cm to metres:

Start with what you have: $120 \text{ cm}$
Choose the conversion factor that has "cm" in the denominator (to cancel it) and "m" in the numerator: $\frac{1 \text{ m}}{100 \text{ cm}}$
Multiply:

120 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = \frac{120}{100} \text{ m} = 1.2 \text{ m}

The "cm" units cancel just like numbers do: $\frac{\text{cm}}{\text{cm}} = 1$ .

Tip

Always write the units explicitly. If the units don't cancel correctly, you've used the wrong conversion factor. This catches 90% of conversion mistakes.

The Reverse: Metres to Centimetres

Now convert 1.2 m to cm. This time, you want "cm" to remain and "m" to cancel. Use $\frac{100 \text{ cm}}{1 \text{ m}}$ :

1.2 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 1.2 \times 100 \text{ cm} = 120 \text{ cm}

Notice: when going from a larger unit (m) to a smaller unit (cm), the number gets larger (1.2 → 120). When going from smaller to larger, the number gets smaller (120 → 1.2). This is a useful sanity check.

Common Conversion Factors You'll Use

Quantity	Relationship	Conversion Factors
Length	1 m = 100 cm	$\frac{1 \text{ m}}{100 \text{ cm}}$ , $\frac{100 \text{ cm}}{1 \text{ m}}$
Mass	1 kg = 1000 g	$\frac{1 \text{ kg}}{1000 \text{ g}}$ , $\frac{1000 \text{ g}}{1 \text{ kg}}$
Time	1 h = 60 min	$\frac{1 \text{ h}}{60 \text{ min}}$ , $\frac{60 \text{ min}}{1 \text{ h}}$
Speed	1 km/h = $\frac{1000}{3600}$ m/s	$\frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

Dimensional Analysis is a powerful tool in physics and engineering that lets us check the consistency of equations, derive relationships, and convert units. But why does it work? Let's build the reasoning from the ground up.

1. The Core Idea: Physical Quantities Have Dimensions

Every physical quantity (like length, time, mass) can be expressed in terms of fundamental dimensions. The most common set in mechanics is:

L = Length
M = Mass
T = Time

For example:

Speed has dimensions $[L T^{-1}]$
Force has dimensions $[M L T^{-2}]$
Energy has dimensions $[M L^2 T^{-2}]$

Why this matters: Two quantities can only be meaningfully compared or equated if they have the same dimensions. You cannot add apples to oranges — and you cannot add length to time.

2. The Principle of Dimensional Homogeneity

The key formula that underpins everything is:

Every valid physical equation must be dimensionally homogeneous.

This means: the dimensions on the left-hand side must equal the dimensions on the right-hand side.

Why must this hold?

Consider an equation like:

v = u + at

Left side: $[v] = L T^{-1}$
Right side: $[u] = L T^{-1}$ , $[at] = (L T^{-2})(T) = L T^{-1}$

Both sides have dimensions $L T^{-1}$ . If they didn't match, the equation would be physically meaningless — you'd be comparing quantities that cannot be equal in any real experiment.

Reasoning: Physical laws describe relationships between measurable quantities. If the dimensions don't match, the equation cannot represent a real physical relationship, because the numerical value would depend on the arbitrary choice of units.

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

This is the deeper mathematical reason. The Buckingham Pi Theorem states:

If a physical problem involves $n$ variables and $k$ fundamental dimensions, then it can be reduced to $n-k$ independent dimensionless groups (called $\pi$ groups).

Why does this work?

Imagine you have a relationship:

f(Q_1, Q_2, \dots, Q_n) = 0

where each $Q_i$ has dimensions. Because the equation must be dimensionally homogeneous, we can rearrange it into a function of dimensionless products only:

F(\pi_1, \pi_2, \dots, \pi_{n-k}) = 0

The reasoning: Dimensions act as constraints. Each fundamental dimension (M, L, T) gives one constraint. So if you have $n$ variables and $k$ constraints, you only have $n-k$ independent dimensionless combinations.

Example: For a simple pendulum, the period $T$ depends on length $L$ , mass $m$ , and gravity $g$ . That's 4 variables with 3 dimensions (M, L, T). So $4-3 = 1$ dimensionless group: $\pi = \frac{T^2 g}{L}$ . This tells us $T \propto \sqrt{L/g}$ without solving any differential equation.

4. Why We Can Convert Units Using Dimensional Analysis

The conversion factor formula:

\text{Value in new unit} = \text{Value in old unit} \times \left(\frac{\text{old unit}}{\text{new unit}}\right)^{\text{dimension exponent}}

Why this works: …

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Given data

	Mass of dinitrogen	Mass of dioxygen
(i)	14 g	16 g
(ii)	14 g	32 g
(iii)	28 g	32 g
(iv)	28 g	80 g

The data shows that for a fixed mass of dinitrogen (14 g or 28 g), the masses of dioxygen that combine are in the ratio of small whole numbers. For 14 g of N₂, O₂ masses are 16 g and 32 g — ratio 1:2. For 28 g of N₂, O₂ masses are 32 g and 80 g — ratio 2:5. This obeys the Law of Multiple Proportions.
Note

Statement: When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
Conversions:

Given	Conversion	Result
1 km	to mm	$1 \text{ km} = 10^6 \text{ mm}$
1 km	to pm	$1 \text{ km} = 10^{15} \text{ pm}$
1 mg	to kg	$1 \text{ mg} = 10^{-6} \text{ kg}$
1 mg	to ng	$1 \text{ mg} = 10^6 \text{ ng}$

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The data obey the Law of Multiple Proportions: when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio. Here, fixing 14 g of nitrogen, the oxygen masses are 16 g, 32 g, 32 g, and 40 g — ratios 1 : 2 : 2 : 2.5 (or 2 : 4 : 4 : 5). The unit conversions are: (i) 1 km = 10⁶ mm = 10¹⁵ pm;

(ii) 1 mg = 10⁻⁶ kg = 10⁶ ng;

(iii) 1 mL = 10⁻³ L = 10⁻³ dm³.

Part (a) — The Law of Chemical Combination

1. What the data is telling us

We have four experiments where nitrogen and oxygen react. Look at the masses:

Experiment	N₂ (g)	O₂ (g)
(i)	14	16
(ii)	14	32
(iii)	28	32
(iv)	28	80

The key insight: the mass of nitrogen is not the same in all cases. To test the Law of Multiple Proportions, we must fix the mass of one element and see how the mass of the other varies.

2. Fix the mass of nitrogen

Take 14 g of nitrogen as the reference.

In (i), oxygen is already 16 g.
In (ii), oxygen is 32 g.
In (iii), we have 28 g of nitrogen — that’s twice 14 g. So the oxygen that would combine with 14 g of nitrogen is half of 32 g = 16 g.
In (iv), again 28 g of nitrogen → half of 80 g = 40 g of oxygen per 14 g N₂.

So, for a fixed 14 g of nitrogen, the masses of oxygen are:

16 g, 32 g, 16 g, 40 g

3. Find the ratios

Divide each by the smallest (16 g):

16 ÷ 16 = 1
32 ÷ 16 = 2
16 ÷ 16 = 1
40 ÷ 16 = 2.5

These are 1 : 2 : 1 : 2.5. Multiply through by 2 to clear the decimal: 2 : 4 : 2 : 5 — a simple whole-number ratio.

Watch out

A common mistake is to compare the oxygen masses directly without first fixing the nitrogen mass. If you just look at the raw numbers, you might think the ratios are 16 : 32 : 32 : 80 = 1 : 2 : 2 : 5 — which is also a simple ratio, but that’s coincidental. The law requires fixing one element’s mass. Always do that step.

4. Which law is this?

This is the Law of Multiple Proportions (Dalton, 1803). It states:

Law of Multiple Proportions : When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

Here, nitrogen and oxygen form several oxides (NO, NO₂, N₂O₃, N₂O₅, etc.). The data matches exactly.

Tip

You can also check by looking at the compounds formed:

14 g N + 16 g O → NO (molar mass 30 g, N:O = 14:16)
14 g N + 32 g O → NO₂ (N:O = 14:32)
28 g N + 32 g O → N₂O₂? Actually that’s 2 × NO, so same ratio as (i).
28 g N + 80 g O → N₂O₅ (since 2×14 g N + 5×16 g O = 28 g + 80 g). The oxygen masses per fixed nitrogen are 16, 32, 16, 40 — exactly the masses needed for NO, NO₂, NO, and N₂O₅.

Part (b) — Unit Conversions

1. 1 km = ? mm = ? pm

We go stepwise:

1 km = 1000 m = 10³ m …

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Common Mistakes Students Make on This Question (Unit Conversion & Law of Multiple Proportions)

Part (a): Law of Multiple Proportions

Mistake 1: Confusing the law with Law of Definite Proportions

What students do wrong: They think the fixed mass ratio of N:O in each compound proves the Law of Definite Proportions.
Why it's wrong: The Law of Definite Proportions says a given compound always has the same ratio. Here, we have different compounds (different ratios), so it's the Law of Multiple Proportions.
How to avoid: Ask yourself: Are we comparing different compounds or the same compound? If different ratios exist for the same elements, it's Multiple Proportions.

Mistake 2: Not calculating the ratio correctly

What students do wrong: They compare masses directly without fixing one element's mass.
Why it's wrong: The law requires comparing masses of one element that combine with a fixed mass of the other.
How to avoid:
- Fix the mass of nitrogen (say 14 g).
- Find the masses of oxygen that combine with it:
  - (i) 14 g N + 16 g O → ratio = 16
  - (ii) 14 g N + 32 g O → ratio = 32
  - (iii) 28 g N + 32 g O → for 14 g N, O = 16 g → ratio = 16
  - (iv) 28 g N + 80 g O → for 14 g N, O = 40 g → ratio = 40
- The oxygen masses (16, 32, 16, 40) are in simple whole-number ratios: 1:2:1:2.5 → but 2.5 = 5/2, so multiply by 2 → 2:4:2:5 → simple whole numbers.

Mistake 3: Forgetting the statement of the law

What students do wrong: They write a vague or incomplete statement.
How to avoid: Memorise the exact statement:

"If two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers."

Part (b): Unit Conversions

Mistake 4: Mixing up prefixes (kilo, milli, nano, pico)

What students do wrong: They confuse the order or magnitude of prefixes.
How to avoid: Memorise the prefix ladder:
- kilo (k) = $10^3$
- milli (m) = $10^{-3}$
- nano (n) = $10^{-9}$
- pico (p) = $10^{-12}$
- deci (d) = $10^{-1}$
- centi (c) = $10^{-2}$

Mistake 5: Incorrect conversion factor for km → mm

What students do wrong: They think 1 km = $10^3$ m = $10^3 \times 10^3$ mm = $10^6$ mm (wrong!).
Why it's wrong: 1 m = 1000 mm = $10^3$ mm, so 1 km = $10^3$ m × $10^3$ mm/m = $10^6$ mm.
How to avoid: Write step-by-step:
- $1 \text{ km} = 10^3 \text{ m}$
- $1 \text{ m} = 10^3 \text{ mm}$
- So $1 \text{ km} = 10^3 \times 10^3 = 10^6 \text{ mm}$

Mistake 6: Confusing pm (picometer) with nm (nanometer)

What students do wrong: They use $10^{-9}$ for pico instead of $10^{-12}$ .
How to avoid: Remember: pico = $10^{-12}$ , nano = $10^{-9}$ $1 0^{- 9}$ .
- $1 \text{ km} = 10^3 \text{ m} = 10^3 \times 10^{12} \text{ pm} = 10^{15} \text{ pm}$

Mistake 7: mg → kg conversion error

What students do wrong: They think 1 mg = $10^{-3}$ kg (wrong!).
Why it's wrong: 1 mg = $10^{-3}$ g, and 1 g = $10^{-3}$ kg, so 1 mg = $10^{-3} \times 10^{-3} = 10^{-6}$ kg.
How to avoid: Use the chain:
- $1 \text{ mg} = 10^{-3} \text{ g}$
- $1 \text{ g} = 10^{-3} \text{ kg}$
- So $1 \text{ mg} = 10^{-3} \times 10^{-3} = 10^{-6} \text{ kg}$ …

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Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

The Intuition: Same Quantity, Different Labels

The Precise Statement

How to Convert: The Only Rule You Need

The Reverse: Metres to Centimetres

Common Conversion Factors You'll Use

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

1. The Core Idea: Physical Quantities Have Dimensions

2. The Principle of Dimensional Homogeneity

Why must this hold?

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

Why does this work?

4. Why We Can Convert Units Using Dimensional Analysis

Why this works: …

Part (a) — The Law of Chemical Combination

1. What the data is telling us

2. Fix the mass of nitrogen

3. Find the ratios

4. Which law is this?

Part (b) — Unit Conversions

1. 1 km = ? mm = ? pm

Solution Method: Law of Multiple Proportions (with Unit Conversion)

Part (a) — Identifying the Law

Part (b) — Unit Conversions

(i) 1 km = ? mm = ? pm

(ii) 1 mg = ? kg = ? ng

Common Mistakes Students Make on This Question (Unit Conversion & Law of Multiple Proportions)

Part (a): Law of Multiple Proportions

Mistake 1: Confusing the law with Law of Definite Proportions

Mistake 2: Not calculating the ratio correctly

Mistake 3: Forgetting the statement of the law

Part (b): Unit Conversions

Mistake 4: Mixing up prefixes (kilo, milli, nano, pico)

Mistake 5: Incorrect conversion factor for km → mm

Mistake 6: Confusing pm (picometer) with nm (nanometer)

Mistake 7: mg → kg conversion error

Part (a) — The Law of Chemical Combination

1. What the data is telling us

2. Fix the mass of nitrogen

3. Find the ratios

4. Which law is this?

Part (b) — Unit Conversions

1. 1 km = ? mm = ? pm

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